What is two times the energy stored in the capacitor (in μJ ) between terminals ' a ' and ' b ' of the network (as shown in figure)?If capacitance of each capacitor is 1 μF .

Step-by-step solution

Redrawing network, we get a balanced Wheatstone's network. Charge on capacitor between terminals a and b Q 2 = C V 2 Energy stored in that capacitor = 1 2 Q 2 2 C = Q 2 8 C = CV 2 8 = 1 × 10 - 6 × 10 2 8 J = 12 . 5 μ J